Published by Patrick Mutisya · 14 days ago
Know that a conductor moving across a magnetic field or a changing magnetic field linking with a conductor can induce an e.m.f. in the conductor.
| Quantity | Expression |
|---|---|
| Induced e.m.f. (\$\mathcal{E}\$) | \$\mathcal{E} = -\frac{d\Phi}{dt}\$ |
| Magnetic flux (\$\Phi\$) | \$\Phi = B A \cos\theta\$ |
Where \$B\$ is the magnetic field strength (T), \$A\$ the area of the loop (m²), and \$\theta\$ the angle between \$B\$ and the normal to the loop.
When a straight conductor of length \$L\$ moves with velocity \$v\$ perpendicular to a uniform magnetic field \$B\$, the induced e.m.f. is:
\$\mathcal{E} = B L v \sin\theta\$
\$\theta\$ is the angle between the direction of motion and the magnetic field. Maximum e.m.f. occurs when \$\theta = 90^\circ\$ (motion ⟂ field).
For a coil of \$N\$ turns, if the magnetic field through the coil changes, the induced e.m.f. is:
\$\mathcal{E} = -N\frac{d\Phi}{dt} = -N\frac{d}{dt}(B A \cos\theta)\$
Any change in \$B\$, \$A\$, or \$\theta\$ will produce an e.m.f.
| Factor | Effect on \$\mathcal{E}\$ |
|---|---|
| Magnetic field strength (\$B\$) | Directly proportional – stronger \$B\$ gives larger \$\mathcal{E}\$. |
| Length of conductor in field (\$L\$) | Directly proportional – longer conductor cuts more field lines. |
| Speed of motion (\$v\$) | Directly proportional – faster motion increases rate of flux change. |
| Number of turns (\$N\$) in a coil | Directly proportional – each turn contributes to total e.m.f. |
| Angle between motion and field (\$\theta\$) | Proportional to \$\sin\theta\$ – maximum when \$\theta = 90^\circ\$. |
| Rate of change of magnetic field (\$dB/dt\$) | Directly proportional – rapid change gives larger \$\mathcal{E}\$. |
The induced e.m.f. produces a current whose magnetic field opposes the original change in flux. This can be determined using the right‑hand rule for generators:
A rod 0.15 m long moves at 4.0 m s⁻¹ through a magnetic field of 0.25 T directed into the page. Find the magnitude of the induced e.m.f.
Solution:
\$\mathcal{E}=BLv = (0.25)(0.15)(4.0)=0.15\ \text{V}\$
A single loop of wire of area 0.02 m² lies in a uniform magnetic field that increases from 0.30 T to 0.50 T in 0.10 s. Calculate the average induced e.m.f.
Solution:
\$\Delta\Phi = BfA - BiA = (0.50-0.30)(0.02)=0.004\ \text{Wb}\$
\$\mathcal{E}_{\text{avg}} = -\frac{\Delta\Phi}{\Delta t}= -\frac{0.004}{0.10}= -0.040\ \text{V}\$
Magnitude = 0.040 V.
A coil of 200 turns has a radius of 5 cm and is placed in a magnetic field that is decreasing at a rate of \$2.0\times10^{-3}\ \text{T s}^{-1}\$. Find the induced e.m.f.
Solution:
\$A = \pi r^2 = \pi(0.05)^2 = 7.85\times10^{-3}\ \text{m}^2\$
\$\mathcal{E}= -N A \frac{dB}{dt}= -(200)(7.85\times10^{-3})(-2.0\times10^{-3})=3.14\times10^{-3}\ \text{V}\$